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1.
By Vadim2, 8 years ago, In Russian
min-cost-max-flow(проблема решена) Добрый день сообщество CF. Я начал изучать потоки и столкнулся с проблемами. Недавно я нашёл алгоритм максимального потока минимальной стоимости. Алгоритм 1) Для каждого ребра, соединяющего вершины u и v добавим обратное, соединяющее вершины v и u, с пропускной способностью равной нулю и стоимостью, равной минус стоимости прямого ребра. 2) Введем для каждой вершины u потенциал φu и расстояние от источника du. Пусть сначала они будут равны нулю. 3) Для каждого ребра e (в том числе и для обратного), соединяющего вершины u и v, введем вес weight(e) = cost(e) + φu — φv (именно для определения веса и нужны потенциалы — без них вес мог бы быть отрицательным для обратных ребер и нельзя было бы применить алгоритм Дейкстры), где cost(e) — стоимость ребра. 4) С помощью алгоритма Дейкстры находим расстояния от источника до всех вершин (в поиске участвуют только те ребра, поток по которым не максимальный). Если нет пути от источника до стока, значит максимальный поток н...
min-cost-max-flow(проблема решена), ; g[y].push_back(z); } int maxflow; /// max поток int mincost; /// min цена int main, ].push_back(z); } int maxflow; /// max поток int mincost; /// min цена int main() { int

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2.
By dario2994, 11 months ago, In English
Editorial of Global Round 11 #### General comments Broadly speaking, problems A-B-C-D were "div2 problems", while F-G-H were "strong grandmaster problems" (with E staying in the middle). I did not expect anyone to solve all the problems and thus I decided to give the scoring F+G=H (so that maybe someone would have solved H). <br><br> Many of the problems (A, C, D, E, G) admit multiple solutions. Sometimes the core of the solution is the same (C, D) and sometimes the solutions are truly different (A, E, G). <br><br> If you are an experienced participant, I would like to hear your opinion on the problems. Feel free to comment on this post or send me a private message. <br><br> <spoiler summary="Overview of the problemset" > The easiest problem of the contest, *A-Avoiding Zero*, is about rearranging an array of numbers. It is intended as a very easy problem that still requires to think. Then, in *B-Chess Cheater* an intuitive (but nontrivial to prove) greedy approach is the way to go. *C-The Hard Work of P...
algorithm to compute the min-cost maxflow and optimize Dijkstra as much as possible (and simplify the graph, the shortest-augmenting-path algorithm to compute the min-cost maxflow and optimize Dijkstra as much

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3.
By AFGhazy, history, 6 years ago, In English
http://codeforces.com/gym/100534/problem/I [Coin Robbery] some help Hey Codeforces, I have question in this problem my code run as follow: - run Dijkstra to get the shortest path - run Dijkstra again to see if any edge get to the destination with cost > shortest_path then cancel this edge - run max flow with vertex split why this approach is wrong ?! thanks in advance. this is my code : ~~~~~ #include <bits/stdc++.h> using namespace std; const int INF = 1000 * 1000 * 1000; int n,m; int g[105][105], cost; int dijkstra(int source){ vector<int > dist(105, INF); dist[source] = 0; set<pair<int, int > > q; q.insert(make_pair(0, source)); while(!q.empty()){ int current = q.begin()->second; int distance = q.begin()->first; q.erase(q.begin()); if(current == n-1) return distance; for(size_t i = 0; i < n; ++i) { if(g[current][i] + distance < dist[i]){ if(q.count(make_pair(dist[i], i))) q.erase(make_pair(dist[i], i)); ...
,g[a][b]); g[b][a] = min(c,g[b][a]); } cost = dijkstra(0); dijkstrA(0, ][a] = min(c,g[b][a]); } cost = dijkstra(0); dijkstrA(0); build(); cost

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4.
By majk, history, 2 years ago, In English
Codeforces Global Round 4 Editorial [tutorial:1178A] <spoiler summary="Code"> ``` #include <iostream> #include <vector> using namespace std; int main() { int N; cin >> N; vector<int> A(N); for (int&a:A) cin >> a; vector<int> P{1}; int rest = 0, cur = A[0]; for (int i = 1; i < N; ++i) { if (A[i] <= A[0]/2) { cur += A[i]; P.push_back(i+1); } else { rest += A[i]; } } if (cur > rest) { cout << P.size() << endl; for (int i = 0; i < P.size(); ++i) cout << P[i] << " \n"[i==P.size()-1]; } else { cout << 0 << endl; } } ``` </spoiler> [tutorial:1178B] <spoiler summary="Code"> ``` #include <iostream> #include <string> using namespace std; typedef long long ll; int main() { string S; cin >> S; ll a = 0, b = 0, c = 0; for (int i = 0; i < S.size(); ++i) { if (S[i] == 'o') { b += a; } else if (i > 0 && S[i-1] == 'v')...
}; for (int i = 0; i < l; ++i) lo = min(lo, {C[a+i],i}); int j = lo.second

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5.
By tomasf, history, 4 years ago, In English
2017 USP-ICMC Editorial #### [A &mdash; Bath Temperature](http://codeforces.com/gym/101484/problem/A) **Problem author:** [user:teochaban,2017-08-18] <spoiler summary="Solution"> **Editorial author:** [user:tomasf,2017-08-29] If any tap has the desired temperature, there is no need to open a second tap, so the answer is 1. Since the resulting temperature is a weighted mean of the opened taps, all obtainable temperatures are between the minimum and maximum temperature available on the taps. If the desired temperature is outside that range, the answer is -1. Otherwise, we can use the minimum and maximum temperature to obtain the desired temperature, so the answer is 2. **Overall Complexity:** $O(N)$ **Code([user:tomasf,2017-08-22]):** [Link](https://pastebin.com/dmSX2L39) </spoiler> #### [B &mdash; Nicoleta's Cleaning](http://codeforces.com/gym/101484/problem/B) **Problem author:** [user:tomasf,2017-08-18] <spoiler summary="Solution"> **Editorial author:** [user:tomasf,2017-08-29] ...
\le 25 $$ $dp[i][j]$ is the minimum cost of completing the string after having added $i

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Tutorial of 2017 USP-ICMC
 
 
 
 
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6.
By Xellos, 7 years ago, In English
Codeforces Trainings Season 2 Episode 8: Editorial [Complete problemset + tests + original presentation of solutions.](http://www.bapc.eu/problemset.zip) [Official scoreboard](http://bapc.eu/scoreboards/bapc/), [public contest scoreboard](http://www.bapc.eu/scoreboards/public/), [used for training elsewhere](http://www.cs.ubc.ca/~acm-web/practice/2014-10-25/scores.php). Short hints first, more detailed solutions afterwards. ### A. Avoiding the Apocalypse (difficulty: medium-hard, [code](http://ideone.com/JQOMyq)) Maxflow in a suitably selected graph. Ford-Fulkerson is sufficient. <hr> The vertices of our graph need to correspond to states that one person can be in when traversing the original road network; the number of people moving in a group is then represented by flow along edges. Therefore, the most natural choice is to use vertices corresponding to states (location, time). The rules from the problem statement say that at most $c$ people can move along an edge $e=(u->v)$ in the original at any time. That me...
. ### A. Avoiding the Apocalypse (difficulty: medium-hard, [code](http://ideone.com/JQOMyq))Maxflow

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